EDIT: Proving $f^{-1}(f(C))=C$

Question

I need to prove that $f^{-1}(f(C))=C$.

This are the informations. There exists two sets A and B, and function $f(A)\to B$.

I don't know how to solve this, and I tried to search google, but I didn't find anything useful.

Please help. Thanks!!

EDIT: I forgot to add that function is bijective

The equation in the title is false in general. Do you know anything else about $f$? — Michael Albanese, Jan 05 '14 at 15:28
As a counterexample, you can take $A = B = \mathbf{R}$, $f:x \mapsto x^2$, $Z = [0;1]$. — fkraiem, Jan 05 '14 at 15:29
@depecheSoul Do it. If needed, someone will surely help translate or answer directly. — Git Gud, Jan 05 '14 at 15:32
@NasuSama for me also, but I need to prove it and I don't know how. — depecheSoul, Jan 05 '14 at 15:35
@depecheSoul We, the users, want you to show some thoughts for this problem, even if you can't proceed. — NasuSama, Jan 05 '14 at 15:37
Look at Michael Albanese's answer. For this type of problem, when you want to show something is equal, in set theory we show that one is the subset of another, and likewise another is the subset of one (in Layman's terms). — NasuSama, Jan 05 '14 at 15:38
Now that we know that $f$ is bijective, we should ask what $f^{-1}(D)$ means for a subset $D$ of $B$. How is that defined? — Carsten S, Jan 05 '14 at 15:54
For every element $x$ from $A$ it is valid that $y=f(x)\implies x=f^{-1}(y)$, and for $x\in B$, so now I have that $f^{-1}(f(x))=f^{-1}(y)=x$, and now I have to do the same for $x\in B$. Am I on right track. Thanks!! — depecheSoul, Jan 05 '14 at 16:02

score 2 · Answer 1 · answered Jan 05 '14 at 15:37

2

Hint: Show that $f^{-1}(f(C)) \subseteq C$ and $f^{-1}(f(C)) \supseteq C$. One of these is true for any $f$, while the other relies on the fact that $f$ is bijective (in fact, you just need $f$ to be injective).

answered Jan 05 '14 at 15:37

Michael Albanese

99,526

Because $ f^{-1}(f(x))\subset C\ and\ f^{-1}(f(x))\supset C$ is a definition of subsets equality,can I write that $f^{-1}(f(C))=f^{-1}(f(C))$ and go from there??? Thanks – depecheSoul Jan 05 '14 at 15:53
What is $x$ and why are the two subset relationships true? Note that $f^{-1}(f(C))$ and $C$ are two sets. To show $f^{-1}(f(C)) \subseteq C$, you need to show that every element of $f^{-1}(f(C))$ is an element of $C$. – Michael Albanese Jan 05 '14 at 15:59
I tried to solve it something like this: For every element $x$ from $A$ it is valid that $y=f(x)\implies x=f^{-1}(y)$, and for $x\in B$, so now I have that $f^{-1}(f(x))=f^{-1}(y)=x$, and now I have to do the same for $x\in B$. Am I on right track. Thanks for helping me!! – depecheSoul Jan 05 '14 at 16:06
First of all, that implication requires justification. Once you have justified it, you have shown that $f^{-1}(f(x)) = x$. Which of the two subset relationships does this help you to prove? – Michael Albanese Jan 05 '14 at 16:10
What do you mean by justification? – depecheSoul Jan 05 '14 at 16:20
I mean that you should say $y = f(x)$ implies $x = f^{-1}(y)$ because $f$ is a bijection. – Michael Albanese Jan 06 '14 at 12:32

score 1 · Accepted Answer · answered Jan 05 '14 at 21:09

Here is a somewhat bigger hint, in a different style than you may be used to.

First, which elements $\;x\;$ does the set $\;f^{-1}[f[C]]\;$ contain? Let's use the basic properties and calculate: \begin{align} & x \in f^{-1}[f[C]] \\ \equiv & \;\;\;\;\;\text{"basic property of $\;\cdot^{-1}[\cdot]\;$: $\;x \in f^{-1}[W] \;\equiv\; f(x) \in W\;$"} \\ & f(x) \in f[C] \\ \equiv & \;\;\;\;\;\text{"basic property of $\;\cdot[\cdot]\;$: $\;y \in f[V] \;\equiv\; \langle \exists x : x \in V : f(x) = y \rangle\;$"} \\ & \langle \exists z : z \in C : f(z) = f(x) \rangle \\ \text{...} & \;\;\;\;\;\text{"..."} \\ \end{align}

Now we want to end this calculation with $\;x \in C\;$ (why?), and there are two ways to continue it: \begin{align} & \langle \exists z : z \in C : f(z) = f(x) \rangle \\ \Leftarrow & \;\;\;\;\;\text{"choose a specific $\;z\;$ that gets us near our goal..."} \\ & \text{...} \\ \end{align} and also \begin{align} & \langle \exists z : z \in C : f(z) = f(x) \rangle \\ \Rightarrow & \;\;\;\;\;\text{"assume $\;f\;$ has some specific property to achieve our goal..."} \\ & \text{...} \\ \end{align}

How do you complete the calculations? What property of $\;f\;$ do you need? What is your conclusion?

EDIT: Proving $f^{-1}(f(C))=C$

2 Answers2